Lower Secondary Science
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Verysuperkiasu:
Yes the amount of work needed to lift the object at constant velocity will be the same as the potential energy that the ball possesses. And if the upward force is greater than the weight, then the ball will accelerate upwards and gain kinetic energy in addition to the GPE.
In other words, can we say that the change in potential energy is the same as the work done on the object?Dr.Daniel:
A force of 10N acting over a distance of 5m would raise the ball at constant velocity and now the ball would have 50J of gravitational potential energy.
Thinking in terms of the law of conservation of energy, the work done on the ball goes into an increase of gravitational potential energy. If I used a greater force and input more work, then the ball would gain kinetic energy in addition to the gravitational potential energy.
Actually I came across this question (below):
A rock lifted up to the top of the cliff 120m from the ground has a potential energy of 12kJ. How much work has been done to lift the rock?
a) 0J b)10kJ c)12kJ d)1440kJ
Can't use formula workdone = force x dist moved as we don't know the force applied. And since the answer is c)12kJ, I thought maybe P.E. is the same as work done (afterall they're both in Joules)...that's what led me to raise this question in the first place. -
sunkc2000:
This is a common hurdle that students have to go through when they make the transition from primary to secondary - How do you explain science concepts in writing? I make two general suggestions..My dd just gotten back her Sec 2 term 1 test paper. Realised she did badly for her open ended questions. I read through her papers and found that she, probably has the knowledge, but phrasing the answers were a challenge to her. Hence, she was not able to score full marks for most for the questions.
Can anyone advise how can she improve in that area?
Does teacher look at Key words (like Pri school) when marking the papers?
1. Practice is important. You need to practice expressing thoughts verbally as well as in writing. The more you practice speaking and writing science explanations the better you will get.
2. Reading is important also because by reading, you get the hang of how to express ideas in writing.
I usually discourage students from attempting to memorize answers to structured questions or simply trying to put certain keywords in an answer. The keyword association can sometimes result in the student not really understanding, just writing certain keywords. And this leads to trouble as a student progresses through secondary school. It is being able to apply a concept to a specific question - and explain the connection clearly - that is the important skill to develop. -
Does anybody know a good tuition centre for IP Secondary 3 students?
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Hi Pasta Greely,
You may want to check out this topic thread yourself \"Secondary School Science\": http://www.kiasuparents.com/kiasu/forum ... =29&t=7748
You can see Dr. Daniel's replies to a lot of questions on Science and/or Physics questions by parents.
I have been following this topic thread to help my children. When any one of my children wants to take Science or Physics tuition, I will definitely go to Dr. Daniel coz I find his explanation very clear and easy to understand. -
I have no idea what to do... Should I use v=ir formulae but idk how http://i61.tinypic.com/k0sh3r.jpg
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Michaelia0816:
I have no idea what to do... Should I use v=ir formulae but idk how http://i61.tinypic.com/k0sh3r.jpg
The voltage across YZ is the same as the voltage across WX. Both resistors have the same amount of current going through them. Current is never destroyed in circuits with resistors. Both these resistors are in the main path, so if they have the same resistance and same current, then using V=IR, they have the same voltage.
Voltage across YZ is 3.6 V
Voltage across XY must be the remaining voltage 9V – 3.6 – 3.6 = 1.8V
Also if you look at the equivalent resistance of the three resistors in parallel – it is 0.5 Ohms. You get that by taking 1/Re = ½ + ½ +1/1. So the same current going through half the resistance gives you half the voltage or 1.8 V.
The voltage across each 2 Ohm resistor is the same = 1.8V. For circuit elements in parallel, they have the same voltage across them.
The current at X is 3.6A. The same current that goes into the first resistor comes out of the first resistor.
The current at point Z is the same also = 3.6A
For the current through each 2 Ohm resistor, use V=IR. The voltage across the 2 Ohm resistor is 1.8V.
1.8V = I (2 Ohms)
I = 0.9Amps. -
Can someone help me with this physics question.... :thankyou:
![](\"http://i57.tinypic.com/a4t0y0.jpg\")
http://i57.tinypic.com/a4t0y0.jpg\">
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![](\"http://i62.tinypic.com/2yv2a34.jpg\")
![](\"http://i57.tinypic.com/i3xvo7.jpg\")
Please help me with this question...Thanks -
Mary Joy:
Not too much information given here, so I'm going to make the assumption that the M113 has to be afloat in pure water. In order for a body to float in a liquid, it needs to have a lower density as compared to the liquid. In the case of pure water, its density is 1000 kg/m3.Can someone help me with this physics question.... :thankyou:
Since density = mass / volume,
1000 = 1.115 X 10^4 / volume
Hence volume = 11.15 m3
So the M113 would need a volume greater than that to stay afloat.
Hope this helps!
P.S The print is a little small so the value I used for the mass of the M113 could be wrong, but the principle is still the same. -
Mary Joy:
According to the graph, when the light intensity is zero then the number of oxygen bubbles is zero. Since oxygen is a byproduct of photosynthesis, this shows that light is essential for photosynthesis to occur.
Please help me with this question...Thanks
What is not 100% clear from the question is whether there was actually a data point taken with 0 light intensity. It seems so since both graphs are extended down to 0 light intensity. To cover any uncertainty in this, you could add.....
The entire apparatus could be covered to ensure zero light intensity and then the volume of oxygen collected could be measured before and after to confirm that no oxygen is produced in the dark.
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